2a. A wave has a frequency of 50 Hz and a wavelength of 0.10 m. How do you find the wave speed? *

The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21

frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is $h(t) = 3+21.5\cdot t-4.9\cdot t^{2}$ after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 seconds and after 4 seconds, (b) When does the projectile reach its maximum height? (c) What is the maximum height? (d) When does it hit the ground? (e) With what velocity does it hits the ground?

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

$v(t) = 21.5-9.8\cdot t$ (Eq. 1)

Where $v(t)$ is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

$v(2) = 21.5-9.8\cdot (2)$

$v(2) = 1.9\,\frac{m}{s}$

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

$v(4) = 21.5-9.8\cdot (4)$

$v(4) = -17.7\,\frac{m}{s}$

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for $t$ :

$21.5-9.81\cdot t = 0$

$t = 2.192\,s$

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

$h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}$

$h (2.192) = 26.584\,m$

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for $t$ . That is:

$3+21.5\cdot t-4.9\cdot t^{2} = 0$

Roots are found by means of Quadratic Formula:

$t_{1}\approx 4.523\,s$ and $t_{2}\approx -0.135\,s$

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

$v(4.523) = 21.5-9.81\cdot (4.523)$

$v(4.523) = -22.871\,\frac{m}{s}$

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

5 0

4 years ago

A star is estimated to be 4.8 km away from the earth. When we see this star in the night sky how old is the image. Let the speed

andre [41]

Time = distance / speed

Time = (4,800 meters) / (3 x 10⁸ m/s)

Time = 0.000016 second

This number is not one of the choices on the list. My hunch is that you copied the distance wrong.

If the estimated distance to the star is actually 4.8 x 10¹⁵ km, instead of 4.8 km, then the answer would be close to 500 years (B).

There's no way a star can be "4.8 km away from the Earth". You can walk that far in about an hour, and passenger jet airplanes fly twice as far as that away from the Earth !

5 0

3 years ago

What happens to a current when the electric field in the wire increase

Gnom [1K]

Answer:

If the voltage is increased then the electric field is higher, and electron velocity (average) is proportional to this field. Then you have an increase in speed. And current is total charge passing per time unit, so current is proportional to velocity value of charge (and to voltage in resistors and wire).

Explanation:

5 0

4 years ago

1. A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts

Snezhnost [94]

Answer:

397 j

Explanation:

Because 5.0kg yuh

3 0

2 years ago

A wire of arbitrary shape, which is confined to the x-y plane, carries a current i from point a to point b in the plane. show th

SSSSS [86.1K]

Answer:

See explanation

Explanation:

Solution:-

- A wire of arbitrary shape,which is confined to the x-y plane,carries a current I from point A to point B in the x-y plane.

- See diagram (attached) for clarity.

- Let’s assume that the horizontal distance between A and B is "s" and the vertical distance between A and B is "d". Then for the straight line path vector ( L ):

L = s i^ + d j^

- The force on the straight wire with current I is then:

F = I * ( L x B )

Where, L: The path vector between points A and B

B: The magnetic field strength vector

For the curved wire vector "ds = dx i^ + dy j^" and the force on the wire is:

F = ∫ [ I (ds x B) = I ∫ (dx i^ + dy j^) x B

When current "I" and magnetic field "B" are uniform then we can pull both of them out of the integral. Separate the integral and calculate each differential separately:

F = I ∫ (dx i^) x B + I ∫ (dy j^) x B

= I (s i^ x B) + I ( d j^ x B ) = I ( L x B )

- The force of curved and straight line have the same force:

F = I ( L x B ) acting on them.

4 0

3 years ago